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  • I need help with - Complex numbers and quadratic equations - JEE Main-10

The value of 'a' for which sum of squares of roots of x^{2}-\left ( a-2 \right )x-\left ( a+1 \right )= 0  is minimum is

  • Option 1)

    -1

  • Option 2)

    0

  • Option 3)

    1

  • Option 4)

    2

 

Answers (1)

best_answer

Let \alpha ,\, \beta are roots of given equation.

\alpha ^{2}+\beta ^{2}=\left ( \alpha +\beta \right )^{2}-2\alpha \beta =\left ( a-2 \right )^{2}+2\left ( a+1 \right )

\Rightarrow \: \alpha ^{2}+\beta ^{2}=a^{2}-2a+6= \left ( a-1 \right )^{2}+5

\Rightarrow \: \alpha ^{2}+\beta ^{2}  will be minimum when \left ( a-1 \right )^{2}=0

\Rightarrow \: a=1

\therefore Option (C)

 

Product of Roots in Quadratic Equation -

\alpha \beta = \frac{c}{a}

- wherein

\alpha \: and\ \beta are roots of quadratic equation:

ax^{2}+bx+c=0

a,b,c\in C

 

 


Option 1)

-1

This is incorrect

Option 2)

0

This is incorrect

Option 3)

1

This is correct

Option 4)

2

This is incorrect

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